2021 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

This question tests the limit and the differentiability of a piecewise function at its breakpoint. First check continuity by comparing the limit with the value at the point, and then compute the derivative at the point from the difference quotient; the two results together determine the correct option.

Steps:

1. Continuity at $x=0$: $\lim\limits_{x\to0} \frac{e^x-1}{x}=1=f(0)$, so $f$ is continuous at $0$; this excludes the options claiming a maximum or a minimum, since the function is given by one and the same formula on both sides of $0$ and therefore has no extremum there.
2. Differentiability at $x=0$: $\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x-0}=\lim\limits_{x\to0}\dfrac{e^x-1 - x}{x^2}=\tfrac{1}{2}$, so the derivative exists and is not $0$, which rules out (C) and leaves (D).

Answer:

D.

Approach:

This question tests the use of the chain rule along two different curves in order to determine the partial derivatives of a function of two variables, and then the construction of the total differential from them. The two identities hold on two curves; evaluating them at suitable points and solving the resulting linear system gives $f_x(1,1)$ and $f_y(1,1)$.

Steps:

1. In (1) take $x=0,y=0$: the arguments become $x+1=1$ and $e^y=1$, and the right-hand side is $(0+1)^2+2\cdot0\cdot(0+1)=1$, so $f_x(1,1)+f_y(1,1)=1$.
2. In (2) take $x=1$: the arguments become $x=1$ and $x^2=1$, and the right-hand side is $4\cdot1\ln1+2\cdot1=2$, so $f_x(1,1)+2f_y(1,1)=2$.
3. Solving these two linear equations gives $f_x(1,1)=0,\ f_y(1,1)=1$.
4. Hence $df(1,1)=f_x(1,1)dx+f_y(1,1)dy=dy$, which is option (C).

Answer:

C.

Approach:

This question tests the composition of Taylor expansions. The function is a product of two factors, so both factors are expanded to the required order and the product is then formed, keeping the terms up to $x^3$.

Steps:

1. Expand the first factor: $\sin x = x-\dfrac{x^3}{6}+o(x^3)$.
2. Expand the second factor: $\dfrac{1}{1+x^2}=1 - x^2 + o(x^2)$.
3. Multiply and keep the terms up to $x^3$: $f(x)=\left(x-\dfrac{x^3}{6}+o(x^3)\right)\left(1-x^2+o(x^2)\right)=x-\dfrac{x^3}{6}-x^3+o(x^3)=x-\dfrac{7}{6}x^3+o(x^3)$.
4. Comparing this with $ax+bx^2+cx^3$ gives $a=1,\ b=0,\ c=-\dfrac{7}{6}$, so (A) is the correct choice.

Answer:

A.

Approach:

This question tests the definition of the definite integral as a limit of Riemann sums. Divide $[0,1]$ into $n$ equal subintervals and check, option by option, whether the number of sample points, the length of each subinterval and the step size are consistent with each other.

Steps:

1. The definition of the definite integral as a Riemann sum: divide $[0,1]$ into $n$ equal subintervals, each of length $\dfrac1n$, and take the midpoint $\xi_k=\dfrac{2k-1}{2n}$ of each subinterval; then $\displaystyle\int_0^1f(x)dx=\lim_{n\to\infty}\sum_{k=1}^nf\!\left(\dfrac{2k-1}{2n}\right)\dfrac1n$ (this holds when $f$ is continuous on $[0,1]$).
2. Check the options one by one: in (A) there are $n$ sample points but the step size is $\dfrac{1}{2n}$, so the total length is $\dfrac12$ and it is not $\int_0^1f(x)dx$; in (C) there are $2n$ sample points but the step size is $\dfrac1n$, so the total length is $2$, which is likewise wrong.
3. In option (D) the sample points $\dfrac{k}{2n}$ are the right endpoints rather than the midpoints (the first one is $0$, an endpoint), and the step size is $\dfrac2n$, so the total length is $2$; this is not the required integral either.
4. Only in option (B) are the sample points the midpoints, the number of them is $n$ and the step size is $\dfrac1n$, so the three match the Riemann sum of $\int_0^1f(x)dx$ exactly, and (B) is the correct choice.

Answer:

B.

Approach:

This question tests the index of inertia of a quadratic form. First expand the quadratic form and write it in matrix form, then compute the eigenvalues; the signs of the eigenvalues give the positive and the negative index of inertia.

Steps:

1. Expand the quadratic form: $f(x_1,x_2,x_3)=(x_1+x_2)^2+(x_2+x_3)^2-(x_3-x_1)^2=2x_2^2+2x_1x_2+2x_2x_3+2x_1x_3$, so the matrix of the quadratic form is $A=\begin{pmatrix}0&1&1\\1&2&1\\1&1&0\end{pmatrix}$.
2. Compute the characteristic polynomial: $|\lambda E-A|=\lambda(\lambda-3)(\lambda+1)$, so the eigenvalues are $\lambda_1=3,\ \lambda_2=0,\ \lambda_3=-1$.
3. The values above are confirmed by $\operatorname{tr}A=2$, $|A|=0$ and $\operatorname{tr}A^2=10$.
4. There is only one positive eigenvalue, namely $3$, and only one negative eigenvalue, namely $-1$, so the positive index of inertia is 1 and the negative index of inertia is 1, and (B) is the correct choice.

Answer:

B.

Approach:

This question tests the Gram-Schmidt orthogonalisation of a system of vectors. The orthogonalisation formula is used to remove, in turn, the components of each vector along the directions of the vectors already obtained.

Steps:

1. First $\beta_1=\alpha_1=(1,0,0)^T$ is already normalised, and the coefficient that makes $\beta_2$ orthogonal to it is $k=\dfrac{[\alpha_2,\beta_1]}{[\beta_1,\beta_1]}=2$, so $\beta_2=\alpha_2-2\alpha_1=(0,2,0)^T$.
2. The coefficient that makes $\beta_3$ orthogonal to $\beta_1$ is $l_1=\dfrac{[\alpha_3,\beta_1]}{[\beta_1,\beta_1]}=\dfrac{5}{2}$.
3. The coefficient that makes $\beta_3$ orthogonal to $\beta_2$ is $l_2=\dfrac{[\alpha_3,\beta_2]}{[\beta_2,\beta_2]}=\dfrac{1}{2}$.
4. With these values the three vectors are pairwise orthogonal, so $l_1=\dfrac{5}{2}$ and $l_2=\dfrac{1}{2}$, and (A) is the correct choice.

Answer:

A.

Approach:

This question tests the rank of block matrices. The rank of a block matrix $\begin{pmatrix}A&0\\ C&A^T\end{pmatrix}$ is the dimension obtained by superimposing the row (column) spaces of $A$ and of $C$ and then superimposing the row space of $A^T$, so each option has to be examined separately.

Steps:

1. For option (A): the block matrix is $\begin{pmatrix}A&0\\0&A^TA\end{pmatrix}$, the two blocks are independent of each other, and the rank is $r(A)+r(A^TA)=2r(A)$, so it holds.
2. For option (B): $\begin{pmatrix}A&0\\AB&A^T\end{pmatrix}$; the row vectors of $AB$ are linear combinations of the row vectors of $A$ and add no new dimension to the row space, so the rank is $r(A)+r(A^T)=2r(A)$, and it holds.
3. For option (C): $\begin{pmatrix}A&0\\BA&A^T\end{pmatrix}$; the rows of the lower block $BA$ are in general not linear combinations of the rows of $A$ (they combine the rows of $B$ with the columns of $A$) and may bring an extra dimension to the row space, making the rank greater than $2r(A)$, so the equality does not always hold.
4. For option (D): $\begin{pmatrix}A&B\\0&A^T\end{pmatrix}$; in the same way the rows of $B$ may bring an extra dimension to the row space, so this equality does not always hold either. The statement that does not hold is option (C).

Answer:

C.

Approach:

This question tests the monotonic relations between conditional probabilities. Each statement is checked by expanding the conditional probabilities with the law of total probability, and a counterexample is looked for where the relation fails.

Steps:

1. By the law of total probability $P(A|A\cup B)=\dfrac{P(A)}{P(A)+P(B)-P(AB)}$ and $P(\bar A|A\cup B)=\dfrac{P(A\cup B)-P(A)}{P(A)+P(B)-P(AB)}$, so the condition $P(A|A\cup B)\geq P(\bar A|A\cup B)$ only says $P(A)\geq P(A\cup B)-P(A)$, that is $2P(A)\geq P(A\cup B)-P(A)$, and it does not force $P(A)>P(B)$.
2. For instance, taking $P(A)=P(B)=\tfrac12$ and $P(AB)=0$ gives $P(A|A\cup B)=P(\bar A|A\cup B)=\tfrac12$, so the hypothesis of (D) holds while $P(A)=P(B)$, and (D) does not hold.
3. For (A): $P(A|B)=P(A)$ means that $A$ and $B$ are independent, and then $P(A|\bar B)=P(A)$ as well, so (A) holds.
4. For (B) and (C): by the law of total probability $P(A)=P(A|B)P(B)+P(A|\bar B)(1-P(B))$, so $P(A|B)>P(A)$ forces $P(A|\bar B)>P(A)$, and $P(A|B)>P(A|\bar B)$ forces $P(A|B)>P(A)$; hence (B) and (C) hold, and the statement that does not hold is (D).

Answer:

D.

Approach:

This question tests the unbiasedness of a linear combination of sample means and the computation of its variance. The expectation is obtained from the linearity of the expectation, and the variance is obtained from the covariance between the two sample means.

Steps:

1. By the linearity of the expectation, $E(\hat\theta)=E(\bar X)-E(\bar Y)=\mu_1-\mu_2=\theta$, so $\hat\theta$ is an unbiased estimator of $\theta$.
2. The variance of a difference of two correlated sample means is $D(\bar X-\bar Y)=D(\bar X)+D(\bar Y)-2\operatorname{Cov}(\bar X,\bar Y)$.
3. Here $D(\bar X)=\dfrac{\sigma_1^2}{n}$, $D(\bar Y)=\dfrac{\sigma_2^2}{n}$, and $\operatorname{Cov}(\bar X,\bar Y)=\dfrac{\operatorname{Cov}(X_1,Y_1)}{n}=\dfrac{\rho\sigma_1\sigma_2}{n}$.
4. Hence $D(\hat\theta)=\dfrac{\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2}{n}$, so (C) is the correct choice.

Answer:

C.

Approach:

This question tests the probability of a type II error in a hypothesis test. The sample mean is standardised with the help of the normal distribution, and the probability of not rejecting the null hypothesis when it is false is then read off from the standard normal distribution function.

Steps:

1. Since $X_1,\dots,X_{16}$ are independent and follow $N(\mu,4)$, the sample mean satisfies $\bar X\sim N\!\left(\mu,\tfrac{1}{4}\right)$, so its standard deviation is $\tfrac12$.
2. When $\mu=11.5$, the rejection region is $\{\bar X\geq11\}$, so a type II error occurs exactly when $\bar X<11$.
3. Therefore $P(\bar X<11)=P\!\left(Z<\dfrac{11-11.5}{1/2}\right)=P(Z<-1)=1-\Phi(1)$, so (B) is the correct choice.

Answer:

B.

Approach:

This question tests the evaluation of an improper integral by completing the square. The quadratic $x^2+2x+2$ is rewritten as $(x+1)^2+1$, after which the integral becomes the antiderivative of $\dfrac{1}{1+(x+1)^2}$, that is, an arctangent, evaluated between the two limits.

Steps:

1. Complete the square: $x^2+2x+2=(x+1)^2+1$.
2. With the substitution $u=x+1$ the lower limit $x=0$ becomes $u=1$, while the improper upper limit stays infinite.
3. The antiderivative is $\arctan u$, so $\int_0^{+\infty}\dfrac{dx}{(x+1)^2+1}=\left.\arctan(x+1)\right|_0^{+\infty}=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$.

Answer:

$\dfrac{\pi}{4}$.

Approach:

This question tests the second derivative of a function given by parametric equations. First $\dfrac{dy}{dx}$ is computed from the ratios of the derivatives with respect to $t$, and then the result is differentiated once more with respect to $t$ and divided by $\dfrac{dx}{dt}$.

Steps:

1. From $x=2e^t+t+1$ and $y=4(t-1)e^t+t^2$ we get $\dfrac{dx}{dt}=2e^t+1$ and $\dfrac{dy}{dt}=4e^t+2t$.
2. Hence $\dfrac{dy}{dx}=\dfrac{4e^t+2t}{2e^t+1}$.
3. Differentiating once more with respect to $t$ and dividing by $\dfrac{dx}{dt}$ gives $\dfrac{d^2y}{dx^2}=\dfrac{(4e^t+4e^t+2)(2e^t+1)-(4e^t+2t)2e^t}{(2e^t+1)^3}$.
4. The condition $x=0$ corresponds to $t=0$, and substituting $t=0$ gives $\dfrac{2}{3}$.

Answer:

$\dfrac{2}{3}$.

Approach:

This question tests the solution of a Cauchy-Euler equation. Substituting a trial solution $y=x^\lambda$ gives the characteristic equation, whose roots determine the general solution; the initial conditions then fix the two constants.

Steps:

1. Put $y=x^\lambda$; then $y'=\lambda x^{\lambda-1}$ and $y''=\lambda(\lambda-1)x^{\lambda-2}$, and the equation $x^2y''+y'-4y=0$ becomes $\lambda(\lambda-1)x^\lambda+\lambda x^{\lambda-1}-4x^\lambda=0$, that is $\lambda^2-4=0$.
2. The roots are $\lambda=2$ and $\lambda=-2$, so the general solution is $y=C_1 x^2+C_2 x^{-2}$.
3. The initial values give $y(1)=C_1+C_2=1,\ y'(x)=2C_1 x-2C_2 x^{-3}$, and $y'(1)=2C_1-2C_2=2$.
4. Solving the two equations gives $C_1=1,C_2=0$, so $y=x^2$.

Answer:

$y=x^2$.

Approach:

This question tests the application of the divergence theorem to a surface integral. The original integral is a second-kind surface integral over the whole outer surface of the closed region $\{(x,y,z)\mid x^2+4y^2\le4,\ 0\le z\le2\}$, so it can be turned directly into a triple integral.

Steps:

1. Write the integral as the flux of a vector field: $F=(x^2,\ y^2,\ z)$, and $\Sigma$ is the whole outer surface of the given region (the lateral side together with the upper and the lower cap).
2. By the divergence theorem, $\displaystyle\iint_{\Sigma}x^2\,dy\,dz+y^2\,dz\,dx+z\,dx\,dy=\iiint_{\Omega}\operatorname{div}F\,dV$, where $\operatorname{div}F=2x+2y+1$.
3. The region $\Omega=\{(x,y,z)\mid x^2+4y^2\le4,\ 0\le z\le2\}$ is symmetric about both the $x$-axis and the $y$-axis, so $\iiint_\Omega2x\,dV=0$ and $\iiint_\Omega2y\,dV=0$.
4. The integral is therefore $\iiint_\Omega1\,dV$, that is, the volume. The cross-section is an ellipse with semiaxes $2$ and $1$, of area $\pi\cdot2\cdot1=2\pi$, and the height is $2$, so the volume is $2\pi\cdot2=4\pi$.

Answer:

$4\pi$.

Approach:

This question tests the relation between a determinant and the cofactors of a matrix. Since the entries of every row of $A$ sum to $2$, the vector $(1,1,1)^T$ is an eigenvector of $A$ with eigenvalue $2$, and the adjugate can then be written in terms of that eigenvector.

Steps:

1. The condition on the row sums gives $A\alpha=\lambda\alpha$ with $\alpha=(1,1,1)^T$ and $\lambda=2$.
2. The adjugate satisfies $A^*=\dfrac{|A|}{\lambda}\alpha\alpha^T$, and the sum of the entries of the first column of $A^*$ is exactly $A_{11}+A_{21}+A_{31}$.
3. That sum is $\dfrac{|A|}{\lambda}\cdot(1+1+1)\cdot1=\dfrac{|A|}{\lambda}$, and with $|A|=3$ and $\lambda=2$ this gives $A_{11}+A_{21}+A_{31}=\dfrac{|A|}{\lambda}= \dfrac{3}{2}$.

Answer:

$\dfrac{3}{2}$.

Approach:

This question tests the computation of the correlation coefficient of two discrete random variables. The joint distribution of $(X,Y)$ is listed first, and the covariance and the two variances are then computed from it.

Steps:

1. The joint distribution is $(X,Y)\sim \{(0,0),(0,1),(1,0),(1,1)\}$ with probabilities $3/10,1/5,1/5,3/10$ respectively.
2. Hence $EX=EY=\dfrac12$, so $DX=DY=\dfrac14$.
3. Moreover $E(XY)=\dfrac{3}{10}$ and $EX\cdot EY=\dfrac14$, so $\mathrm{cov}(X,Y)=E(XY)-EX\cdot EY=\dfrac{3}{10}-\dfrac14=\dfrac1{20}$.
4. Therefore the correlation coefficient is $\rho_{XY}=\dfrac{1/20}{\sqrt{1/4\cdot1/4}}=\dfrac{1}{5}$.

Answer:

$\dfrac{1}{5}$.

Approach:

This question tests the use of a small-order expansion to evaluate a limit whose numerator contains an integral with a variable upper limit. The numerator is expanded up to the order $x^2$, and its two contributions are then combined.

Steps:

1. For the integral, since $e^x-1\sim x$ and $e^t\sim1$ on $[0,x]$, we have $\int_0^x \frac{t^2 e^t}{e^x-1}dt\sim\int_0^x t^2dt=\frac{x^3}{3}+o(x^3)$.
2. Also $\sin x = x-\frac{x^3}{6}+o(x^3)$, so $\dfrac{1}{\sin x}=\dfrac{1}{x}\left(1+\frac{x^2}{6}+o(x^2)\right)$.
3. The numerator is therefore $1+\frac{x^3}{3}+o(x^3)-\dfrac{1}{\sin x}=\frac{1}{2}x^2+o(x^2)$.
4. Dividing by $x^2$ and taking the limit gives $\dfrac{1}{2}$.

Answer:

$\dfrac{1}{2}$.

Approach:

The factor $e^{-x}$ does not affect the convergence, so one first studies the power series $\sum_{n=1}^{\infty}\dfrac{x^{n+1}}{n(n+1)}$ and then sums it with the help of $\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$.

Steps:

1. The ratio test gives the radius $R=1$.
2. At $x=1$ the series $\sum\dfrac{1}{n(n+1)}$ converges, and at $x=-1$ it converges absolutely, so the domain of convergence is $[-1,1]$.
3. For $|x|<1$ we have

4. Substituting the endpoints into the original series gives

Answer:

Approach:

This question tests the maximum distance from the points of a space curve to a coordinate plane. First $z$ is eliminated by means of the equation of the plane, so that the curve is projected onto the $xOy$ plane as an ellipse, and then the extrema of $|z|$ are found with the Lagrange multiplier method.

Steps:

1. From the second equation $z=30-4x-2y$, so the distance to the $xOy$ plane is $|z|=|30-4x-2y|$.
2. Substituting $z=30-4x-2y$ into the first equation $x^2+2y^2-z=6$ gives $x^2+4x+2y^2+2y-36=0$, that is $(x+2)^2+2\left(y+\tfrac12\right)^2=\tfrac{81}{2}$, which shows that the projection of the curve onto the $xOy$ plane is an ellipse.
3. Applying the Lagrange multiplier method to $|z|$ gives the extremal points $(4,1,12)$ and $(-8,-2,66)$, and substituting back into $z=30-4x-2y$ gives $z=12$ and $z=66$ respectively.
4. Comparing the values of $|z|$ at the two ends, the maximum distance from the points of the curve to the $xOy$ plane is $66$.

Answer:

$66$.

Approach:

The integrand of the area integral is symmetric about the origin, so the maximum corresponds to the disc $x^2+y^2\le 4$; the curve integral is then evaluated with Green’s theorem.

Steps:

1. The maximum of $I(D)$ is attained when $x^2+y^2\le4$, and then $I(D_1)=\int_0^{2\pi}\int_0^{2}(4-r^2)r\,dr\,d\theta=8\pi$.
2. Along $\partial D_1$, Green’s theorem turns the curve integral into an area integral, whose value is $-\pi$.

Answer:

$I(D_1)=8\pi$, and the curve integral is $-\pi$.

Approach:

The eigenvalues of $A$ are $\lambda_1=a+2,\ \lambda_2=\lambda_3=a-1$, and the corresponding orthogonal eigenvectors are used to build $P$; one then sets $C=P\Lambda^{1/2}P^T$.

Steps:

1. The characteristic polynomial of $A$ factors as $(\lambda-a-2)(\lambda-a+1)^2$, so the eigenvalues are $a+2$ and $a-1$ (double), and the orthogonal eigenvectors give the matrix $P$ below.
2. One may take

and then $P^T A P=\operatorname{diag}(a+2,a-1,a-1)$. 3. Using $C=P\Lambda^{1/2}P^T$ one obtains

Answer:

$P$ and $C$ are as above.